Question

What is the expression for $$K_{sp}$$ of $$A{g}_{2}Cr{O}_4$$ ?

Answer 1

Hint: Silver chromate is an inorganic ionic salt formed from two silver cations and a chromate anion. The solubility product expression for this salt can be written by writing the chemical equation of the equilibrium between solid silver chromate and the dissociated ions.

Complete answer:
The solubility product $$K_{sp}$$ is a modified equilibrium constant calculated at a particular temperature. In order to write the expression for solubility product, the dissociation reaction of silver chromate must be known.
Even though silver is a transition metal that can show variable oxidation states, it shows $$+1$$ oxidation state in silver chromate salt. The chromate anion carries a net negative charge of $$-2$$ . In order to balance the charges so as to achieve overall neutrality, each molecule of silver chromate contains two silver ions bonded to one chromate ion. Hence, on dissociation an equilibrium is established between the undissociated silver chromate and the ions present in aqueous medium. The equilibrium expression can be written as follows:
$$A{g}_{2}Cr{O}_4(s)\rightleftharpoons 2A{g}^{+}(aq)+Cr{O_4}^{2-}(aq)$$
The solubility product expression can be written as the product of concentrations of the aqueous ions (that have been dissociated) raised to the power of their stoichiometric coefficients.
Hence, the expression for $$K_{sp}$$ is :
$$K_{sp}=[A{g}^{+}{]}^2[Cr{O_4}^{2-}]$$

Note:
The solubility product of silver chromate is very low and therefore even at low concentration of the ions in water, the product of concentrations exceed the solubility product value resulting in an easy precipitation of silver chromate salt that has a characteristic brick red colour due to charge transfer.